Something tells me this must be a fairly simple question, but I have somehow been unable to find an answer to it. In short: I need to calculate the difference between two signals, A and B, each one of them with its own signal-to-noise ratio. What is the SNR of the resulting value?

As for the context: we are doing differential photometry, generating the light curve of some celestial bodies by comparing their instrumental magnitudes (a logarithmic measure of its brightness) to those of others. These magnitudes are derived from the flux of each star, which is also used in order to estimate the signal-to-noise ratio, given by the following formula [Astronomical Photometry, A. Henden, 1982]:

$$
SNR = \frac{(star\;counts - sky\;counts)}{\sqrt{star\;counts}}
$$

Note that, in case the background were neligible, the SNR could be obtained as $star\;counts / {\sqrt{star\;counts}}$, as for photon arrivals the statistical noise fluctuation is represented by the Poisson distribution, according to Henden. Also, counts are a measurement of the flux, as we are using a CCD camera in order to conduct photometry.

SNR to error (in magnitudes)

A S/N of 100 means that the noise causes the counts to fluctuate about the mean by an amount equal to one hundredth of the mean value. To compute this error in magnitudes, we compare the mean number of counts, $c$, to the maximum or minimum values induced by noise, that is

1 Answer
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Treating the signals as time series: If the first signal $S_1$ has a noise component $N_1$ added to it, then the noisy signal is $S_1+N_1$, similarly the second signal is $S_2+N_2$, so the difference signal would be $(S_1+N_1)-(S_2+N_2)$ and its signal to noise ratio would be $\langle(S_1-S_2)^2\rangle\over\langle(N_1-N_2)^2\rangle$

If the signals are uncorrelated $\langle(S_1-S_2)^2\rangle$ is just $\langle S_1^2\rangle+\langle S_1^2\rangle$. If the signals are correlated, you will have to estimate the covariance $\langle S_1S_2\rangle$ to compute $\langle S_1^2\rangle+\langle S_1^2\rangle-2\langle S_1S_2\rangle$

Thanks a whole lot, twistor59. Although your answer is exactly what I had been looking for, I am not quite sure how it could be applied in our case. I have updated my question in order to include the formula that we use, given the SNR, to compute the error in magnitudes. The thing is that our data set does not distinguishes between signal and noise components — we have a magnitude, say 8.02, and an reliable estimation of its SNR, say 300. [Continues in next comment]
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plokJan 28 '12 at 22:19

From this, and using Henden's formula, I can obtain the error in magnitudes: $-2.5 \log {1\pm\frac{1}{300}} = 0.0036$. Hopefully I'm not wrong when I consider this error (in Henden's words) to be our noise component. It is at this point where I'm stuck. In order to use your formula, would the signal, $S_1^2$, be still 8.02?
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plokJan 28 '12 at 22:20

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Firstly a disclaimer - I'm not familiar with astronomical measurements, just general signal processing. That said, I'm guessing you want to do something like this - you have raw measurements for two stars A and B and you want to compute the SNR on the difference between these measurements. If so, then the formulas I was talking about would only make sense if the measurements of the 2 stars were made at the same time. So I have two series $A_1, A_2,...A_N$, and $B_1,B_2,...B_N$ made at times $t_1,t_2,...t_N$. You then need a method to compute the noise (error) on series A (continued...)
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twistor59Jan 30 '12 at 8:01

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and the noise on series B. In the formula I described, you could either (1) have a value of the noise component for each measurement, i.e. two time series $N_{A1},N_{A2},..$, $N_{B1},N_{B2},...$ and compute the variance explicitly, or (2)have some assumptions on the noise (like uncorrelatedness and Gaussianity) where variance of the combined noise signal would be the sum of the two variances. The other thing I'd like to add is that you need to be careful whether you're treating the signals in the log domain or the linear domain when you compute variances.
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twistor59Jan 30 '12 at 8:12